Answer:
A) F=-20.16×10⁹N
B) if the distance doubles, force is 4 times smaller.
Explanation:
q1=-28C
q2=5mC=0.005C
d=25cm=0.25m
Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.
Thus:
F=9×10⁹×(-28)×0.005/0.25²
F=-20.16×10⁹N
The minus sign indicates attraction.
If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.
Answer:
Explanation:
Let the linear charge density of the charged wire is given as
here we can use Gauss law to find the electric field at a distance r from wire
so here we will assume a Gaussian surface of cylinder shape around the wire
so we have
here we have
so we have
<span>37.8 seconds
First, determine the speed difference between the car and truck.
95 km/h - 75 km/h = 20 km/h
Convert that speed into m/s to make a more convenient unit of measure.
20 km/h * 1000 m/km / 3600 s/h = 5.556 m/s
Now it's simply a matter of dividing the distance between the two vehicles and their relative speed.
210 m / 5.556 m/s = 37.8 s
So it will take 37.8 seconds for the car to catch the truck that's 210 meters in front of the car.</span>
Answer:
b. equal in size and opposite in direction
Answer:
calculate by urself u don't know this question
